3.1270 \(\int \frac {e^{-3 \tanh ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx\)

Optimal. Leaf size=221 \[ -\frac {2 a^2 \sqrt {c-a^2 c x^2}}{x^2 \sqrt {1-a^2 x^2}}-\frac {\sqrt {c-a^2 c x^2}}{4 x^4 \sqrt {1-a^2 x^2}}+\frac {a \sqrt {c-a^2 c x^2}}{x^3 \sqrt {1-a^2 x^2}}+\frac {4 a^4 \log (x) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}-\frac {4 a^4 \sqrt {c-a^2 c x^2} \log (a x+1)}{\sqrt {1-a^2 x^2}}+\frac {4 a^3 \sqrt {c-a^2 c x^2}}{x \sqrt {1-a^2 x^2}} \]

[Out]

-1/4*(-a^2*c*x^2+c)^(1/2)/x^4/(-a^2*x^2+1)^(1/2)+a*(-a^2*c*x^2+c)^(1/2)/x^3/(-a^2*x^2+1)^(1/2)-2*a^2*(-a^2*c*x
^2+c)^(1/2)/x^2/(-a^2*x^2+1)^(1/2)+4*a^3*(-a^2*c*x^2+c)^(1/2)/x/(-a^2*x^2+1)^(1/2)+4*a^4*ln(x)*(-a^2*c*x^2+c)^
(1/2)/(-a^2*x^2+1)^(1/2)-4*a^4*ln(a*x+1)*(-a^2*c*x^2+c)^(1/2)/(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6153, 6150, 88} \[ \frac {4 a^3 \sqrt {c-a^2 c x^2}}{x \sqrt {1-a^2 x^2}}-\frac {2 a^2 \sqrt {c-a^2 c x^2}}{x^2 \sqrt {1-a^2 x^2}}+\frac {a \sqrt {c-a^2 c x^2}}{x^3 \sqrt {1-a^2 x^2}}-\frac {\sqrt {c-a^2 c x^2}}{4 x^4 \sqrt {1-a^2 x^2}}+\frac {4 a^4 \log (x) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}-\frac {4 a^4 \sqrt {c-a^2 c x^2} \log (a x+1)}{\sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c - a^2*c*x^2]/(E^(3*ArcTanh[a*x])*x^5),x]

[Out]

-Sqrt[c - a^2*c*x^2]/(4*x^4*Sqrt[1 - a^2*x^2]) + (a*Sqrt[c - a^2*c*x^2])/(x^3*Sqrt[1 - a^2*x^2]) - (2*a^2*Sqrt
[c - a^2*c*x^2])/(x^2*Sqrt[1 - a^2*x^2]) + (4*a^3*Sqrt[c - a^2*c*x^2])/(x*Sqrt[1 - a^2*x^2]) + (4*a^4*Sqrt[c -
a^2*c*x^2]*Log[x])/Sqrt[1 - a^2*x^2] - (4*a^4*Sqrt[c - a^2*c*x^2]*Log[1 + a*x])/Sqrt[1 - a^2*x^2]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 6153

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c +
d*x^2)^FracPart[p])/(1 - a^2*x^2)^FracPart[p], Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a,
c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !IntegerQ[n/2]

Rubi steps

\begin {align*} \int \frac {e^{-3 \tanh ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx &=\frac {\sqrt {c-a^2 c x^2} \int \frac {e^{-3 \tanh ^{-1}(a x)} \sqrt {1-a^2 x^2}}{x^5} \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\sqrt {c-a^2 c x^2} \int \frac {(1-a x)^2}{x^5 (1+a x)} \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\sqrt {c-a^2 c x^2} \int \left (\frac {1}{x^5}-\frac {3 a}{x^4}+\frac {4 a^2}{x^3}-\frac {4 a^3}{x^2}+\frac {4 a^4}{x}-\frac {4 a^5}{1+a x}\right ) \, dx}{\sqrt {1-a^2 x^2}}\\ &=-\frac {\sqrt {c-a^2 c x^2}}{4 x^4 \sqrt {1-a^2 x^2}}+\frac {a \sqrt {c-a^2 c x^2}}{x^3 \sqrt {1-a^2 x^2}}-\frac {2 a^2 \sqrt {c-a^2 c x^2}}{x^2 \sqrt {1-a^2 x^2}}+\frac {4 a^3 \sqrt {c-a^2 c x^2}}{x \sqrt {1-a^2 x^2}}+\frac {4 a^4 \sqrt {c-a^2 c x^2} \log (x)}{\sqrt {1-a^2 x^2}}-\frac {4 a^4 \sqrt {c-a^2 c x^2} \log (1+a x)}{\sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 77, normalized size = 0.35 \[ \frac {\sqrt {c-a^2 c x^2} \left (4 a^4 \log (x)-4 a^4 \log (a x+1)+\frac {4 a^3}{x}-\frac {2 a^2}{x^2}+\frac {a}{x^3}-\frac {1}{4 x^4}\right )}{\sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c - a^2*c*x^2]/(E^(3*ArcTanh[a*x])*x^5),x]

[Out]

(Sqrt[c - a^2*c*x^2]*(-1/4*1/x^4 + a/x^3 - (2*a^2)/x^2 + (4*a^3)/x + 4*a^4*Log[x] - 4*a^4*Log[1 + a*x]))/Sqrt[
1 - a^2*x^2]

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fricas [A]  time = 1.02, size = 503, normalized size = 2.28 \[ \left [\frac {8 \, {\left (a^{6} x^{6} - a^{4} x^{4}\right )} \sqrt {c} \log \left (\frac {4 \, a^{5} c x^{5} + {\left (2 \, a^{6} + 4 \, a^{5} + 6 \, a^{4} + 4 \, a^{3} + a^{2}\right )} c x^{6} + {\left (4 \, a^{4} - 4 \, a^{3} - 6 \, a^{2} - 4 \, a - 1\right )} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, a c x + {\left (4 \, a^{3} x^{3} - {\left (4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} x^{4} + 6 \, a^{2} x^{2} + 4 \, a x + 1\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{4} x^{6} + 2 \, a^{3} x^{5} - 2 \, a x^{3} - x^{2}}\right ) - {\left (16 \, a^{3} x^{3} - {\left (16 \, a^{3} - 8 \, a^{2} + 4 \, a - 1\right )} x^{4} - 8 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{4 \, {\left (a^{2} x^{6} - x^{4}\right )}}, \frac {16 \, {\left (a^{6} x^{6} - a^{4} x^{4}\right )} \sqrt {-c} \arctan \left (-\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, a^{2} + 2 \, a + 1\right )} x^{2} + 2 \, a x + 1\right )} \sqrt {-c}}{2 \, a^{3} c x^{3} - {\left (2 \, a^{3} + a^{2}\right )} c x^{4} + {\left (a^{2} + 2 \, a + 1\right )} c x^{2} - 2 \, a c x - c}\right ) - {\left (16 \, a^{3} x^{3} - {\left (16 \, a^{3} - 8 \, a^{2} + 4 \, a - 1\right )} x^{4} - 8 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{4 \, {\left (a^{2} x^{6} - x^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^5,x, algorithm="fricas")

[Out]

[1/4*(8*(a^6*x^6 - a^4*x^4)*sqrt(c)*log((4*a^5*c*x^5 + (2*a^6 + 4*a^5 + 6*a^4 + 4*a^3 + a^2)*c*x^6 + (4*a^4 -
4*a^3 - 6*a^2 - 4*a - 1)*c*x^4 - 5*a^2*c*x^2 - 4*a*c*x + (4*a^3*x^3 - (4*a^3 + 6*a^2 + 4*a + 1)*x^4 + 6*a^2*x^
2 + 4*a*x + 1)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - c)/(a^4*x^6 + 2*a^3*x^5 - 2*a*x^3 - x^2)) - (
16*a^3*x^3 - (16*a^3 - 8*a^2 + 4*a - 1)*x^4 - 8*a^2*x^2 + 4*a*x - 1)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/
(a^2*x^6 - x^4), 1/4*(16*(a^6*x^6 - a^4*x^4)*sqrt(-c)*arctan(-sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*((2*a^2
+ 2*a + 1)*x^2 + 2*a*x + 1)*sqrt(-c)/(2*a^3*c*x^3 - (2*a^3 + a^2)*c*x^4 + (a^2 + 2*a + 1)*c*x^2 - 2*a*c*x - c)
) - (16*a^3*x^3 - (16*a^3 - 8*a^2 + 4*a - 1)*x^4 - 8*a^2*x^2 + 4*a*x - 1)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 +
1))/(a^2*x^6 - x^4)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} c x^{2} + c} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^5,x, algorithm="giac")

[Out]

integrate(sqrt(-a^2*c*x^2 + c)*(-a^2*x^2 + 1)^(3/2)/((a*x + 1)^3*x^5), x)

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maple [A]  time = 0.05, size = 89, normalized size = 0.40 \[ -\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \sqrt {-a^{2} x^{2}+1}\, \left (16 a^{4} \ln \relax (x ) x^{4}-16 \ln \left (a x +1\right ) x^{4} a^{4}+16 x^{3} a^{3}-8 a^{2} x^{2}+4 a x -1\right )}{4 \left (a^{2} x^{2}-1\right ) x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^5,x)

[Out]

-1/4*(-c*(a^2*x^2-1))^(1/2)*(-a^2*x^2+1)^(1/2)*(16*a^4*ln(x)*x^4-16*ln(a*x+1)*x^4*a^4+16*x^3*a^3-8*a^2*x^2+4*a
*x-1)/(a^2*x^2-1)/x^4

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} c x^{2} + c} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^5,x, algorithm="maxima")

[Out]

integrate(sqrt(-a^2*c*x^2 + c)*(-a^2*x^2 + 1)^(3/2)/((a*x + 1)^3*x^5), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {c-a^2\,c\,x^2}\,{\left (1-a^2\,x^2\right )}^{3/2}}{x^5\,{\left (a\,x+1\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c - a^2*c*x^2)^(1/2)*(1 - a^2*x^2)^(3/2))/(x^5*(a*x + 1)^3),x)

[Out]

int(((c - a^2*c*x^2)^(1/2)*(1 - a^2*x^2)^(3/2))/(x^5*(a*x + 1)^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}{x^{5} \left (a x + 1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*c*x**2+c)**(1/2)/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/x**5,x)

[Out]

Integral((-(a*x - 1)*(a*x + 1))**(3/2)*sqrt(-c*(a*x - 1)*(a*x + 1))/(x**5*(a*x + 1)**3), x)

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